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3 years ago

if the volume of a scuba tank filled with air remains constant and its temperature goes down, what happens to its pressure?

Physics

1 answer:

Vladimir [108]3 years ago

3 0

Answer:

Decreases

Explanation:

Ideal gas law:

PV = nRT

where P is absolute pressure,

V is volume,

n is number of moles,

R is gas constant,

and T is absolute temperature.

If V is constant and T decreases, then P must decrease.

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In Fig.25-46,how much charge is stored on the parallel-plate capacitors by the 12.0 V battery? One is filled with air,and the ot

Fittoniya [83]

Answer:

$Q=7.9\times 10^{-10}\ C$

Explanation:

Given that

V= 12 V

K=3

d= 2 mm

Area=5.00 $ 10#3 m2

Assume that

$ = Multiple sign

# = Negative sign

$A=5\times 10^{-3}\ m^2$

We Capacitance given as

For air

$C_1=\dfrac{\varepsilon _oA}{d}$

$C_1=\dfrac{8.85\times 10^{-12}\times 5\times 10^{-3}}{2\times 10^{-3}}$

$C_1=2.2\times 10^{-11}\ F$

$C_2=\dfrac{K\varepsilon _oA}{d}$

$C_2=\dfrac{3\times 8.85\times 10^{-12}\times 5\times 10^{-3}}{2\times 10^{-3}}\ F$

$C_2=6.6\times 10^{-11}\ F$

Net capacitance

C=C₁+C₂

$C=8.8\times 10^{-11}\ F$

We know that charge Q given as

Q= C V

$Q=12\times 6.6\times 10^{-11}\ C$

$Q=7.9\times 10^{-10}\ C$

6 0

3 years ago

A parallel-plate capacitor is connected to a battery. What happens to the stored energy if the plate separation is doubled while

marta [7]

Answer:

Explanation:

Energy stored in a parallel plate capacitor is given by:

$U=\frac{1}{2}CV^2\\\\C=\frac{A\epsilon_{o}}{d}\\\\then\\\\U=\frac{1}{2}\frac{A\epsilon_{o}}{d}V^2--(1)\\\\$

As capacitor remains connected to the battery so V remains constant. As can be seen from (1) that energy is inversely proportional to the separation between the plates so as 'd' becomes doubled, energy decreases by the factor of 2.

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